for the given volumes
(22.5 + 22.5) * (30.17 - 23.50) * 1.00
... * 4.184
When 22.50 mL of 0.500 M H2SO4 is added to 22.50 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.) (d for water= 1.00 g/mL; c for water =4.184 J/g*C)
2 answers
q rxn = mass H2O x specific heat H2O x (Tfinal-Tinitial)
mass H2O = 22.5 + 22.5 = ?
sp. h. is given
So q = approx 1200 (but that's just an estimate)and it is approx -1200 J since the reaction is exothermic.
That is delta H per 0.0225 mols.
Then dH in J/mol = approx -1200/0.0225 = about -53,000 J/mol. Convert to kJ/mol
mass H2O = 22.5 + 22.5 = ?
sp. h. is given
So q = approx 1200 (but that's just an estimate)and it is approx -1200 J since the reaction is exothermic.
That is delta H per 0.0225 mols.
Then dH in J/mol = approx -1200/0.0225 = about -53,000 J/mol. Convert to kJ/mol
1 answer